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Ch 5 solutions

# X 2 dx 2 2 2 two triangles of height 2 and base 2

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Unformatted text preview: 5.5.1 = − + 8 because (d) −A + A2 = 0 2 Deﬁnition + A2 is possible = 15/22 a constant can move past1 a limit and/or a summation sign. y 248 is 11. Theorem 5.5.4(a) depends on the fact that a constant can move past an integral sign, which by 2 y 11. Theorem 11. 5.5.1 depends onbecause athat a the can move past a past can integraly sign, which by sign, which by 5.5.4(a) Theorem 5.5.4(a) fact constant fact can move limityand/or a summation sign. y Deﬁnition y y is possible thedepends on constantthat a constant an move past an integral y Deﬁnition 5.5.1Deﬁnition 5.5.1 is possible because a constant a limit and/or a summation sign. is possible because a constant can move past can move past limit and/or a summation sign. A2 12. If f (x) ≥ 0 for all x in [a, b] then we know that positivity (or nonnegativity) is preserved under x A2 –5 x – A2 limits ) ≥ 0 for all2 x inalsob(by Deﬁnition 5.5.1) for integrals. nonnegativity) is preserved under and sums, hence [a, ] then we know that positivity – 5 A2 55 x x 12. If f (x (or A – 5 A1 5 A2x inknow] that positivity thatnonnegativity) nonnegativity) is preserved under 12. If f (x) ≥and sums,(x)A≥[a,also (by we [x b then we know (or A 0 1for all hence x then – for a, (or A5 limits 12. If f x in2–1 0 b] x all Deﬁnition 5.5.1) for integrals. positivity 5 1 is preserved under 1 A limits and1 –1 –1 limits 2and sums, Deﬁnition 5.5.1) for integrals. A1 for integrals. alsox (by hence also (by Deﬁnition 5.5.1) A sums, hence 4 4 A1 4 A 1 1 A1 =1 1 (3)(3) = 4 /2 13. (c) A (a) 9 − −A1 + A2 = − + 8 = (b)2. −A = (d) (1)(1 + 2) = −3/2 15/ −A1 + A2 = 0. 2 2 2 1 1 13. (a) A 1 (3)(3) = 9/1 = 2 (b) −A =1− (1)(1 + 2) = −3/2 1 2 13. (a) A = 13. 2 (a)= A/= (3)(3) = 9/2 92 (b) −A = y (b) −A = − 3/2 + 2) = −3/2 − (1)(1 + 2) = − (1)(1 y (3)(3) 2 2 2 22 y y2 21 1 A 1 A A 2 y y A1 y x 1 5x A1 A A1 6 6 6 A2 cx y A2 x c A1 cAx 2 6 A2 cx x 1x 5 15. (a) −A1 + A2 = −5 + 8 = 15/2 5 = 10. A = 2(5) (b)(d) −A1 + A2 = 0 0; A1 = A2 by symmetry. (c) 2 1 (c) −A1 + A2 =1− + 8 = 15/2 (d) −A1 + A2 = 0 1 10 2 (c) −A1 + A2(c)− −A18+ A2 /2 − + 8 = 15/2 (d) −A1 + A2(d)y −A1 + A2 = 0 + = 15= = y= 5 2 2 y 5 y5 y 5 A1 –1 (c) –1 –1 A1 A2 A 32 2 A 22 x 32 3 2 2 –1 2 A1 1 1y A y 1 A x x A2 A1 x A1 + A2 = 1 14. (a) A = (1)(2) = 1 2 1 14. (a) A 1 (1)(2) = 1 1 = 3 2 2 –1 1 51 1 13 (5) + (1) = . 2 22 2 2 1y x A x –1 –11 1A x 1 (d) 1 (b) A = (2)(3/2 + 1/2) = 2 2 1 (b) A 1 (2)(3/2 + 11 2) = 2 = / x 1 A= 1 [π (1)2 ] = π/2. 2 Exercise Set 5.5 139 0 0 (x + 2) dx. f (x) dx = 17. (a) −2 −2 Triangle of height 2 and width 2, above x-axis, so answer is 2. 2 0 (b) f (x) dx = 0 (2 − x) dx. (x + 2) dx + −2 −2 2 Two triangles of height 2 and base 2; answer is 4. |x − 2| dx = 0 6 2 6 (c) (2 − x) dx + 0 (x − 2) dx. 2 Triangle of height 2 and base 2 together with a triangle of height 4 and base 4, so 2 + 8 = 10. −2 6 f (x) dx = (d) (x + 2) dx + −2 −4 −4 6 2 0 (x + 2) dx + 0 (2 − x) dx + 2 (x − 2) dx. Triangle of height 2 and base 2, below axis, plus a triangle of height 2, base 2 above axis, another of height 2 and base 2 above axis, and a triangle of height 4 and base 4, above axis. Thus f (x) = −2 + 2 + 2 + 8 = 10. 19. (a) 0.8 (b) −2.6 2 (c) −1.8 (d) −0.3 2 21. f (x)dx + 2 −1 −1 5 g (x)dx = 5 + 2(−3) = −1. 5 23. f (x)dx = 1 0 3 1 f (x)dx − 0 f (x)dx = 1 − (−2) = 3. 3 25. 4 −1 dx − 5 −1 1 xdx = 4 · 4 − 5(−1/2 + (3 · 3)/2) = −4. 1 27. xdx + 2 0 0 1 − x2 dx = 1/2 + 2(π/4) = (1 + π )/2. 29. False; e.g. f (x) = 1 if x > 0, f (x) = 0 otherwise, then f is integrable on [−1, 1] but not continuous. 31. False; e.g. f (x) = x on [−2, +1]. 33. (a) √ x > 0, 1 − x < 0 on [2, 3] so the integral is negative. (b) 3 − cos x > 0 for all x and x2 ≥ 0 for all x and x2 > 0 for all x > 0 so the integral is positive. 35. If f is continuous on [a, b] then f is integrable on [a, b], and, considering Deﬁnition 5.5.1, for every partition and n choice of f (x∗ ) we have k=1 n m∆xk ≤ k=1 n f (x∗ )∆xk ≤ k k=1 n M ∆xk . This is equivalent to m(b − a) ≤ k=1 f (x∗ )∆xk ≤ k M (b − a), and, taking the limit over max ∆xk → 0 we obtain the result. 10 25 − (x − 5)2 dx = π (5)2 /2 = 25π/2. 37. 0 1 39. (3x + 1)dx = 5/2. 0 41. (a) The graph of the integrand is the horizontal line y = C . At ﬁrst, assume that C > 0. Then the region is a b rectangle of height C whose base extends from x = a to x = b. Thus a C dx = (area of rectangle) = C (b − a). If C ≤ 0 then the rectangle lies below the axis and its integral is the negative area, i.e. −|C |(b − a) = C (b − a). 140 Chapter 5 ber 10, 2008 16:01 ”ISM ET chapter 5” Sheet number 28 Page number 252 black n n ember 2008 16:01 16:01 ”ISM x) ET ,chapter 5” Sheet becomes Page Sheet number Page number∗ )∆ black lim f ( ET C the 5” ( xk ber 10, 10, 2008 (b) Since ”ISM=chapter Riema...
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